Science:Math Exam Resources/Courses/MATH101/April 2015/Question 05 (b)/Solution 1

Since $a_{n}={\frac {1}{(2n+1)^{2}}}$ is positive for any $n\geq 1$, monotonically decreasing, and converges to 0, applying the Alternating Series Estimation Theorem, we have $|S-S_{n}|\leq a_{n+1}={\frac {1}{(2(n+1)+1)^{2}}}={\frac {1}{(2n+3)^{2}}}.$

Now, it is enough to find the smallest $n$ such that ${\frac {1}{(2n+3)^{2}}}\leq {\frac {1}{100}}.$

By taking square roots on both side, the inequality is equivalent with ${\frac {1}{2n+3}}\leq {\frac {1}{10}}$ and hence with $10\leq 2n+3$. Therefore, we can easily obtain the smallest natural number $n$ as 4 satisfying $10\leq 2n+3$.

The answer: $n=\color {blue}4$ and the partial sum is $S_{4}=\color {blue}{\frac {1}{3^{2}}}-{\frac {1}{5^{2}}}+{\frac {1}{7^{2}}}-{\frac {1}{9^{2}}}$